Classical and Quantum Bianchi type I cosmology in K-essence theory Luis O. Pimentel 1, J. Socorro 1,2, Abraham Espinoza-García 2 1 Departamento de Fisica de la Universidad Autonoma Metropolitana Iztapalapa, Mexico 2 Departamento de Física de la DCeI de la Universidad de Guanajuato-Campus León, Mexico lopr@xanu.uam.mx December 12, 2013 Luis O. Pimentel 1, J. Socorro 1,2, Abraham Espinoza-García Class. 2 (UAMI) and Q. k-essence December 12, 2013 1 / 19
Overview Luis O. Pimentel 1, J. Socorro 1,2, Abraham Espinoza-García Class. 2 (UAMI) and Q. k-essence December 12, 2013 2 / 19
Introduction We consider here a simplifyed K-essence model with the lagrangian density of the form, S = d 4 x ] R g 2 + Λ + f(φ) X + L mat, where X = 1 2 µφ µ φ and there is not a self interacting potential for the scalar field. Here L mat is the matter Lagrangian, Λ is the cosmological constant, and g is the determinant of the metric tensor. K-essence was originally proposed as a model for inflation, and then as a model for dark energy, along with explorations of unifying dark energy and dark matter. For details of K-essence applied to dark energy you can see Copeland et al. (2006) Luis O. Pimentel 1, J. Socorro 1,2, Abraham Espinoza-García Class. 2 (UAMI) and Q. k-essence December 12, 2013 3 / 19
Bianchi Type I and Lagrangian The metric of the diagonal Bianchi A class models is ds 2 = (Ndt) 2 + e 2Ω(t) (e 2β(t) ) ij ω i ω j = dτ 2 + e 2Ω(t) (e 2β(t) ) ij ω i ω j, (1) where Ω(t) is a scalar, N the lapse function and β ij = diag(β + + 3β, β + 3β, 2β + ), ω i are one-forms that for the Bianchi type I model are ω 1 = dx 1, ω 2 = dx 2, ω 3 = dx 3. The total Lagrangian density then for this metric becomes (with a fluid ) ( Ω2 L I = e 6 3Ω N β + 2 N β ) ] 2 f(φ) N 2N φ 2 + 2N(ρ + Λ), Luis O. Pimentel 1, J. Socorro 1,2, Abraham Espinoza-García Class. 2 (UAMI) and Q. k-essence December 12, 2013 4 / 19
Hamiltonian using the momenta definition, Π q µ = L q, where q µ = (Ω, β µ +, β, φ), using the energy-momentum conservation the law for a perfect fluid, T µν ;ν = 0, ρ = µ ɛ e 3(ɛ+1)Ω, we assumed an equation of state p = ɛρ, the Hamiltonian density is H = e 3Ω 24 ( Π 2 Ω + 12 f(φ) Π2 φ + Π2 + + Π 2 + b ɛ e 3(ɛ 1)Ω + 48Λe 6Ω), b ɛ = 48µ ɛ. Luis O. Pimentel 1, J. Socorro 1,2, Abraham Espinoza-García Class. 2 (UAMI) and Q. k-essence December 12, 2013 5 / 19
Classical Equations We use the Einstein-Hamilton-Jacobi equation, by making the identification S(Ω,β ±,φ) q = Π µ µ in the Hamiltonian constraint H = 0, which results in ( ) S 2 ( ) S 2 ( ) S 2 12 ( ) S 2 b γ e 3Ω(1 γ) 48Λe 6Ω = 0 Ω β + β f(φ) φ with S = S 1 (Ω) + S 2 (β + ) + S 3 (β ) + S 4 (φ) we have ( ) 2 ds1 ( ) b γ e 3(γ 1)Ω + 48Λe 6Ω 2 + κ Ω = 0 (2) dω ( ) 2 ( ) 2 ds2 κ 2 ds3 + = 0, κ 2 = 0 (3) dβ + dβ ( ) 12 2 ds4 κ 2 φ = 0; κ 2 Ω = κ 2 + + κ 2 + κ 2 φ. (4) f(φ) dφ Here κ φ 2 should have the same sign as f (φ). Luis O. Pimentel 1, J. Socorro 1,2, Abraham Espinoza-García Class. 2 (UAMI) and Q. k-essence December 12, 2013 6 / 19
Classical Solutions The classical Hamilto-Jacobi equation were considered and reduced to quadratures when f (φ) is a power or an exponential, for Λ = 0 and arbitrary ɛ. I show here the particular case of dust ɛ = 0. e 3Ω = 3 4 µ 0τ 2 + β ± = κ ± 3 κ Ln 2 Ω κω 2 4 τ (5) τ. (6) κω 2 4 + 3 4 µ 0τ Here κ Ω 2 = κ + 2 + κ 2 + κ φ 2, and κ φ 2 should have the same sign as f (φ). Luis O. Pimentel 1, J. Socorro 1,2, Abraham Espinoza-García Class. 2 (UAMI) and Q. k-essence December 12, 2013 7 / 19
Continued φ(τ) = (m + 2) { κ 2 φ Exp 2 m Ln 2 m κ 2 φ 48ω Ln 12ω Ln κ 2 φ 3ωκ Ω 2 Ln κ 2 φ 48ω Ln τ κ 2 Ω + 3 4 4 µ 0τ τ κ 2 Ω + 3 4 4 µ 0τ ]} τ κ 2 Ω + 3 4 4 µ 0τ τ κ 2 Ω + 3 4 4 µ 0τ ] ]] 2 m+2, f (φ) = ωφ m,, f (φ) = ωφ 2, ]], f (φ) = ωe mφ,. f (φ) = ω. For Λ 0 solutions were found for ɛ = 1, 0, 1. Luis O. Pimentel 1, J. Socorro 1,2, Abraham Espinoza-García Class. 2 (UAMI) and Q. k-essence December 12, 2013 8 / 19
Isotropization From the Friedmann equation (constraint equation), 3Ω 2 3β 2 + 3β 2 f 4 φ 2 ρ Λ = 0, (7) we can see that isotropization is achieved when the terms with β 2 ± go to zero or are negligible with respect to the other terms in the differential equation. It can be shown that the isotropization can be reached for models that expand forever. A qualitative analysis can be obtained from the classical equation for the volume V = e 3Ω, 16V 2 b ɛ V 1 ɛ 48ΛV 2 = κ Ω 2. (8) That is the equation of motion in the coordinate V of a particle under the potential U(V ) = b ɛ V 1 ɛ 48ΛV 2 with energy E = κ Ω 2. We can now have a qualitative idea of the different solutions from energy diagrams. We assume that b ɛ is non negative. κ Ω 2 is real and could take positive, null or negative values. Luis O. Pimentel 1, J. Socorro 1,2, Abraham Espinoza-García Class. 2 (UAMI) and Q. k-essence December 12, 2013 9 / 19
Isotropization We define an anisotropic density ρ a, that is proportional to the shear scalar, ρ a = β 2 + + β 2, (9) and will compare it with ρ ɛ,ρ φ, and Ω 2. From the Hamilton Jacobi analysis we now that ρ a e 6Ω, ρ φ e 6Ω, Ω 2 48Λ+κ Ω 2 e 6Ω +b ɛ e 3(1+ɛ)Ω (10) and the ratios are ρ a ρ φ constant, ρ a ρ ɛ e 3Ω(ɛ 1), ρ a Ω 2 1 κ Ω 2 + 48Λe 6Ω + b ɛ e 3(1 ɛ)ω. (11) Here we see that for expanding an universe the anisotropic density is dominated by the fluid density (with the exception of the stiff fluid) or by the Ω 2 term and then at late times the isotropization is obtained if the expansion goes to infinity. Hence it is necessary to determine when we have an ever expanding universe. Luis O. Pimentel 1, J. Socorro 1,2, Abraham Espinoza-García Class. 2 (UAMI) and Q. k-essence December 12, 2013 10 / 19
Isotropization 0 Κ 2 0 Κ 2 0 0 Figure: Energy diagram for the volume of the universe. U(V ) = b ɛ V 1 ɛ 48ΛV 2 with energy E = κ Ω 2. From the figure, that is qualitatively correct for ɛ 1, we see that for negative Λ all the expanding solutions will recollapse eventually, regardless of the sign of κ Ω 2. For positive Λ an expanding solution will expand forever. We also note that, when κ Ω 2 < 0(κ Ω 2 = κ + 2 + κ 2 + κ φ 2 ), i.e. when the ghost contribution is dominant, there are contracting solutions that reach a minimum and then expand forever, these solutions do not have a big bang singularity. Luis O. Pimentel 1, J. Socorro 1,2, Abraham Espinoza-García Class. 2 (UAMI) and Q. k-essence December 12, 2013 11 / 19
Quantum Solutions Here by quantum we understand the Wheeeler-DeWitt equation, that in the present case is Ψ + 12 2 Ψ f(φ) φ 2 12s 1 Ψ φ f φ + Q Ψ Ω b ɛe 3(ɛ 1)Ω Ψ 48Λe 6Ω Ψ = 0, (12) where Q and s are any real constants that measure the ambiguity in the factor ordering in the variables Ω and φ and is the three dimensional D Lambertian in the l µ = (Ω, β +, β ) coordinates, with signature (- + +). It was solved by the separation of variables method for the same combinations of Λ, ɛ and f (φ) than the classical solutions and will be published elsewhere. Luis O. Pimentel 1, J. Socorro 1,2, Abraham Espinoza-García Class. 2 (UAMI) and Q. k-essence December 12, 2013 12 / 19
References E.J. Copeland, M. Sami and S. Tsujikawa, Title of the publication Int. J. Mod. Phys. D 15,1753-1936 (2006). Luis O. Pimentel 1, J. Socorro 1,2, Abraham Espinoza-García Class. 2 (UAMI) and Q. k-essence December 12, 2013 13 / 19
The End Luis O. Pimentel 1, J. Socorro 1,2, Abraham Espinoza-García Class. 2 (UAMI) and Q. k-essence December 12, 2013 14 / 19
Isotropization We define an anisotropic density ρ a, that is proportional to the shear scalar, ρ a = β 2 + + β 2, (13) and will compare it with ρ ɛ,ρ φ, and Ω 2. From the Hamilton Jacobi analysis we now that ρ a e 6Ω, ρ φ e 6Ω, Ω 2 48Λ+κ Ω 2 e 6Ω +b ɛ e 3(1+ɛ)Ω (14) and the ratios are ρ a ρ φ constant, ρ a ρ ɛ e 3Ω(ɛ 1), ρ a Ω 2 1 κ Ω 2 + 48Λe 6Ω + b ɛ e 3(1 ɛ)ω. (15) Here we see that for expanding an universe the anisotropic density is dominated by the fluid density (with the exception of the stiff fluid) or by the Ω 2 term and then at late times the isotropization is obtained if the expansion goes to infinity. Hence it is necessary to determine when we have an ever expanding universe. Luis O. Pimentel 1, J. Socorro 1,2, Abraham Espinoza-García Class. 2 (UAMI) and Q. k-essence December 12, 2013 15 / 19
Paragraphs of Text The field equations for the anisotropic cosmological model bianchi type I are the following (the prime is the derivative over the time dτ = Ndt, 3Ω 2 3β + 2 3β 2 f 4 φ 2 ρ Λ = 0, f ( 3Ω φ + φ ) + 1 df 2 dφ φ 2 = 0, (16) 2Ω + 3Ω 2 3Ω β + 3 3Ω β β + + 3β + 2 3β + 3β 2 + f 4 φ 2 + p Λ = 0, (17) 2Ω + 3Ω 2 3Ω β + + 3 3Ω β β + + 3β + 2 + 3β + 3β 2 + f 4 φ 2 + p Λ = 0, (18) 2Ω + 3Ω 2 + 6Ω β + + 2β + + 3β 2 + + 3β 2 + f 4 φ 2 + p Λ = 0 (19) Luis O. Pimentel 1, J. Socorro 1,2, Abraham Espinoza-García Class. 2 (UAMI) and Q. k-essence December 12, 2013 16 / 19
Λ = 0 and ɛ 1 when we consider the time transformations dτ = e 3ɛΩ dt, and the change of variable u = κ Ω 2 + b ɛ e 3(ɛ 1)Ω, this equation for Ω has the solution Ω(T) = Ln θ ɛ T 2 + δ ɛ T ] 1 3(ɛ 1), (20) where θ ɛ = ( ) ɛ 1 2 8 bɛ and δ ɛ = κ 2 ɛ 1 Ω 4. And then the time transformation becomes is τ = (1 ɛ) δ ɛ θɛ T 2 + δ ɛ T ] 1 1 ɛ 2 F 1 (1, 2 ɛ 1 ; ɛ 2 ɛ 1 ; T θ ɛ δ ɛ ), (21) Luis O. Pimentel 1, J. Socorro 1,2, Abraham Espinoza-García Class. 2 (UAMI) and Q. k-essence December 12, 2013 17 / 19
The anisotropy functions and the scalar field are given by β ± = κ ] ± T Ln. (22) 12δ ɛ θ ɛ T + δ ɛ (m + 2) Exp φ(t) = 2 { 2η ω η 1 2ω η m m Ln 2η/ω 1 δ ɛ Ln ]] 2 T m+2 δ ɛ Ln θ ɛt+δ ɛ, f (φ) = ωφ m, m 2 ] } 1 T δ ɛ Ln θ ɛt+δ ɛ, f (φ) = ωφ 2, 1 T 2ω δ ɛ Ln θ ɛt+δ ɛ ]], f (φ) = ωe mφ, m 0 f (φ) = ω ] T θ ɛt+δ ɛ (23) Luis O. Pimentel 1, J. Socorro 1,2, Abraham Espinoza-García Class. 2 (UAMI) and Q. k-essence December 12, 2013 18 / 19
Λ = 0 and ɛ = 1 In this case e 3Ω b2 τ =, (24) 4 with b 2 = κ 2 Ω + 48µ 1 that we assume positive. For the anisotropic functions we have 1 β ± = κ ± 3 Ln( τ), (25) b 2 and the scalar field is given by ] 2 (m + 2) η 4 m+2 2ω b2 Ln( τ), f (φ) = ωφ m, m 2 { } 2η Exp 4 φ(τ) = ω b2 Ln( τ), f (φ) = ωφ 2, ] 2 m m Ln η 4 2ω b2 Ln( τ), f (φ) = ωe mφ, m 0 2η/ω 4 b2 Ln( τ), f (φ) = ω, (26) Luis O. Pimentel 1, J. Socorro 1,2, Abraham Espinoza-García Class. 2 (UAMI) and Q. k-essence December 12, 2013 19 / 19